So Easy!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2350 Accepted Submission(s): 729
Problem Description
A sequence Sn is defined as:
Where a, b, n, m are positive integers.┌x┐is the ceil of x. For example, ┌3.14┐=4. You are to calculate Sn.
You, a top coder, say: So easy!
Input
There are several test cases, each test case in one line contains four positive integers: a, b, n, m. Where 0< a, m < 215, (a-1)2< b < a2, 0 < b, n < 231.The input will finish with the
end of file.
Output
For each the case, output an integer Sn.
Sample Input
2 3 1 2013 2 3 2 2013 2 2 1 2013
Sample Output
4 14 4
#include<iostream> #include<cstring> using namespace std; typedef long long ll; ll a,b,n,m; struct matrix { ll f[2][2]; }; matrix mul(matrix A,matrix B){ matrix s; memset(s.f,0,sizeof s.f); ll i,j,k; for(k=0;k<2;k++) for(i=0;i<2;i++){ if(!A.f[i][k]) continue; for(j=0;j<2;j++){ if(!B.f[k][j]) continue; s.f[i][j]+=A.f[i][k]*B.f[k][j]; s.f[i][j]%=m; } } return s; } matrix pow_mod(matrix A,ll k){ matrix s; s.f[0][0]=s.f[1][1]=1; s.f[0][1]=s.f[1][0]=0; while(k){ if(k&1) s=mul(s,A); A=mul(A,A); k>>=1; } return s; } int main(){ while(cin>>a>>b>>n>>m){ matrix e; ll p=2*a; ll q=b-a*a; e.f[0][0]=p;e.f[0][1]=1; e.f[1][0]=q;e.f[1][1]=0; e=pow_mod(e,n-1); ll ans=((p*e.f[0][0]+2*e.f[1][0])%m+m)%m; cout<<ans<<endl; } return 0; }
时间: 2024-10-05 05:05:17